[c#] 주어진 날짜의 정확한 주 번호를 얻으십시오

정확한 ISO-8601 주 번호를 산출하는 .Net 문화가없는 것 같아서, 부분적으로 정확한 수정을 시도하는 대신 내장 주 결정을 우회하고 수동으로 계산을 수행하고 싶습니다. 결과.

내가 끝내었던 것은 다음 확장 방법입니다.

/// <summary>
/// Converts a date to a week number.
/// ISO 8601 week 1 is the week that contains the first Thursday that year.
/// </summary>
public static int ToIso8601Weeknumber(this DateTime date)
{
    var thursday = date.AddDays(3 - date.DayOfWeek.DayOffset());
    return (thursday.DayOfYear - 1) / 7 + 1;
}

/// <summary>
/// Converts a week number to a date.
/// Note: Week 1 of a year may start in the previous year.
/// ISO 8601 week 1 is the week that contains the first Thursday that year, so
/// if December 28 is a Monday, December 31 is a Thursday,
/// and week 1 starts January 4.
/// If December 28 is a later day in the week, week 1 starts earlier.
/// If December 28 is a Sunday, it is in the same week as Thursday January 1.
/// </summary>
public static DateTime FromIso8601Weeknumber(int weekNumber, int? year = null, DayOfWeek day = DayOfWeek.Monday)
{
    var dec28 = new DateTime((year ?? DateTime.Today.Year) - 1, 12, 28);
    var monday = dec28.AddDays(7 * weekNumber - dec28.DayOfWeek.DayOffset());
    return monday.AddDays(day.DayOffset());
}

/// <summary>
/// Iso8601 weeks start on Monday. This returns 0 for Monday.
/// </summary>
private static int DayOffset(this DayOfWeek weekDay)
{
    return ((int)weekDay + 6) % 7;
}

우선 ((int)date.DayOfWeek + 6) % 7), 요일 번호를 결정합니다 (0 = 월요일, 6 = 일요일).

date.AddDays(-((int)date.DayOfWeek + 6) % 7) 요청 된 주 번호를 선행하는 월요일 날짜를 결정합니다.

3 일 후 목표 목요일이 있으며, 그 주가 몇 년인지 결정합니다.

연도 내 (0부터 시작) 일수를 7 (내림차순)로 나누면 연도에서 0으로 표시되는 주 번호를 얻습니다.

C #에서 정수 계산 결과는 내재적으로 반올림됩니다.

답변

function GetWeekOfYear([datetime] $inputDate)
{
   $day = [System.Globalization.CultureInfo]::InvariantCulture.Calendar.GetDayOfWeek($inputDate)
   if (($day -ge [System.DayOfWeek]::Monday) -and ($day -le [System.DayOfWeek]::Wednesday))
   {
      $inputDate = $inputDate.AddDays(3)
   }

   # Return the week of our adjusted day
   $weekofYear = [System.Globalization.CultureInfo]::InvariantCulture.Calendar.GetWeekOfYear($inputDate, [System.Globalization.CalendarWeekRule]::FirstFourDayWeek, [System.DayOfWeek]::Monday)
   return $weekofYear
}